MCQ
$\mathop {\lim }\limits_{n \to \infty } \frac{{{1^p} + {2^p} + {3^p} + ..... + {n^p}}}{{{n^{p + 1}}}} = $
- ✓$\frac{1}{{p + 1}}$
- B$\frac{1}{{1 - p}}$
- C$\frac{1}{p} - \frac{1}{{p - 1}}$
- D$\frac{1}{{p + 2}}$
✓
Answer
Correct option: A.
$\frac{1}{{p + 1}}$
a
(a) $\mathop {\lim }\limits_{n \to \infty } \frac{{{1^p} + {2^p} + {3^p} + ..... + {n^p}}}{{{n^{p + 1}}}}$
(a) $\mathop {\lim }\limits_{n \to \infty } \frac{{{1^p} + {2^p} + {3^p} + ..... + {n^p}}}{{{n^{p + 1}}}}$
$= \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {\left[ {\frac{{{r^p}}}{{{n^{p + 1}}}}} \right]} $
$= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = 1}^n {{{\left( {\frac{r}{n}} \right)}^p}} = \int_0^1 {{x^p}dx} = \left[ {\frac{{{x^{p + 1}}}}{{p + 1}}} \right]_0^1 = \frac{1}{{p + 1}}$.
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